The angular momentum of each disk individually is not conserved, however the total angular momentum of both disks is conserved because there are no external torques acting.
Start by calculating the total angular momentum of both disks (I'm going to replace "w" by "v" since "w" is confusingly close to "$\omega$"):
$$\begin{align}
L_{total} &= I_a \omega_a + I_b \omega_b \\
&= I.2v + 2I.v \\
&= 4Iv
\end{align}$$
Now we bring the disks into contact and they settle down to a constant speed $\omega_{final}$. The total angular momentum is now:
$$\begin{align}
L_{total} &= I_a \omega_{final} + I_b \omega_{final} \\
&= 3I\omega_{final}
\end{align}$$
Because angular momentum is conserved we just equate our two expressions fo $L_{total}$:
$$ 4Iv = 3I\omega_{final} $$
so:
$$ \omega_{final} = \frac{4}{3}v $$
Now, torque $\times$ time is the angular impulse, and we know that the impulse is equal to the change of momentum. So if we calculate the change in momentum of disk A this is equal to the torque times the time i.e. $Tt$, and we know the initial angular momentum of disk A is $2Iv$ so:
$$\begin{align}
Tt &= I_a 2v - I_a \omega_{final} \\
&= 2Iv - \frac{4}{3}Iv \\
&= \frac{2}{3}Iv
\end{align}$$
And dividing both sides by $t$ gives the answer:
$$ T = \frac{2}{3} \frac{Iv}{t} $$
If there is slipping and friction you now have to deal with an accelerating (non-inertial) frame of reference and axis of rotation.
To use Newton's laws of motion in the non-inertial frame of reference a pseudo-force has to be introduced which acts at the centre of mass, has a magnitude equal to the frictional force and acts in the opposite direction to the frictional force.
That pseudo-force produces a torque about the point of contact with the ground which changes the angular momentum of the disc about that point.
Best Answer
The question implies that when the disc is initially put down it is not rolling without slipping yet. In fact, at $t=0$ it's not yet rolling at all.
Simple conservation of momentum doesn't apply here because friction energy is not conserved.
The disc exerts a force $mg$ (its weight) on the surface, which in turn exerts an equal but opposite Normal force $F_N$, which prevents the ring from penetrating the surface:
$$F_N=mg$$
In the simple friction model a friction force $F_f$ is exerted in the opposite sense of motion, acc.:
$$F_f=\mu F_N=\mu mg,$$
where $\mu$ is the friction coefficient.
We can now set up two equations of motion:
1. Translation:
The force $F_f$ causes acceleration:
$$ma=\mu mg$$
So that:
$$v(t)=\mu gt$$
2. Rotation:
The force $F_f$ also causes a torque $\tau$:
$$\tau=I\alpha,$$
where $\tau=-R \mu mg$, $I=\frac{mR^2}{2}$ and $\alpha=\frac{d\omega}{dt}$, so:
$$-R \mu mg=\frac{mR^2}{2} \frac{d\omega}{dt}$$
$$-2\mu g=R\frac{d\omega}{dt}$$
$$\omega(t)=\omega-\frac{2\mu g}{R}t$$
Rolling without slipping occurs when $v(t)=\omega(t) R$, so with the expressions above:
$$\mu gt=\omega R- 2\mu gt$$
$$t=\frac{\omega R}{3\mu g}$$
Inserting this into $\omega(t)$ we get:
$$\large{\omega(t)=\frac{\omega}{3}}$$
The ring will lose half of its initial rotational momentum before rolling without slipping is achieved.
The initial kinetic energy was:
$$K_0=\frac{mR^2\omega^2}{4}$$
The final kinetic energy $K(t)$, including translational energy is:
$$K(t)=\frac{mR^2}{4}\frac{\omega^2}{9}+\frac{m}{2}\frac{\omega^2 R^2}{9}=\frac{mR^2\omega^2}{12}$$